An initially uncharged air-filled capacitor is connected to a 3.33 V charging source. As a result, 8.13×10−5 C of charge is transferred from one of the capacitor's plates to the other. Then, while the capacitor remains connected to the charging source, a sheet of dielectric material is inserted between its plates, completely filling the space. The dielectric constant κ of this substance is 6.27.
Find the voltage V across the capacitor and
the charge Qf stored by it after the dielectric is
inserted and the circuit has returned to a
steady state.
initially capacitance
C = q/v = 8.13 / 3.33 = 2.441 * 10^-5 F
capcitance after inserting dielectric
C = k Co = 6.27* 2.441* 10^-5
C = 15.308* 10^-5 F
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potential difference even after insertion remains constant = 3.33 V
=====
charge
Q = CV = 15.308* 3.33 = 50.975* 10^-5 C
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